Question: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $46.4$ years; the standard deviation is $9.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living longer than $74.6$ years.
Solution: $46.4$ $37$ $55.8$ $27.6$ $65.2$ $18.2$ $74.6$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $46.4$ years. We know the standard deviation is $9.4$ years, so one standard deviation below the mean is $37$ years and one standard deviation above the mean is $55.8$ years. Two standard deviations below the mean is $27.6$ years and two standard deviations above the mean is $65.2$ years. Three standard deviations below the mean is $18.2$ years and three standard deviations above the mean is $74.6$ years. We are interested in the probability of a bear living longer than $74.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the bears will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $18.2$ years and the other half $({0.15\%})$ will live longer than $74.6$ years. The probability of a particular bear living longer than $74.6$ years is ${0.15\%}$.